Problem: $\overline{BC} = 9$ $\overline{AC} = {?}$ $A$ $C$ $B$ $?$ $9$ $ \sin( \angle BAC ) = \frac{9\sqrt{145} }{145}, \cos( \angle BAC ) = \frac{8\sqrt{145} }{145}, \tan( \angle BAC ) = \dfrac{9}{8}$
Solution: $\overline{BC}$ is the opposite to $\angle BAC$ $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the opposite side and need to solve for the adjacent side so we can use the tan function (TOA) $ \tan( \angle BAC ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\overline{BC}}{\overline{AC}}= \frac{9}{\overline{AC}} $ $ \overline{AC}=\frac{9}{\tan( \angle BAC )} = \frac{9}{\dfrac{9}{8}} = 8$